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Question

# In each of the following find the equation of the hyperbola satisfying the given conditions.: (i)vertices(±2,0),foci(±3,0) (ii)vertices(0,±,5),foci(0,±,8) (iii)vertices(0,±,3),foci(0,±,5) (iv)vertices(±5,0),transverseaxis=8 (v)foci(0,±13),conjugateaxis=24 (vi)foci(±3√5),thelatus−rectum=8 (vii)foci(±,4,0),thelatus−rectum=12 (viii)vertices(0,±6),e=53. (ix)foci(0,±√10),passingthrough(2,3) (x)foci(0,±12),latus−rectum=36

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Solution

## (i) Since,vertics(±2,0) and foci (±,0) lies on X-axis,as coefficient of Y-axis is zero. Hence,equation of hyperbola will be of the form x2a2−y2b2=1 ...(i) Where it is given that a=2 and c=3 ∵c2=a2+b2 ⇒9=4+b2 ⇒b2=5 Put the values of a2=4 and b2=5 in Eq,(i),we get x24−y25=1 (ii) Since,vertics(0,±5) and foci (0,±8) lies on Y-axis,as coefficient of X-axis is zero. Hence,equation of hyperbola will be of the form x2a2−y2b2=1 ...(i) Where it is given that (0,±a)=(0,±5)and foci(0,±c)=(0,±8) a=5 and c=8 ∵c2=a2+b2⇒64=25+b2 ⇒b2=64−25 ⇒b2=39 Put a2=25 and b2=39 in Eq,(i),we get y225−x239=1 (iii) Since,vertics(0,±3) and foci (0,±5) lies on Y-axis,as coordinate of X-axis is zero. Hence,equation of hyperbola will be of the form x2a2−y2b2=1 ...(i) Where it is given that (0,±3)=(0,±a)and foci(0,±5)=(0,±c) a=3 and c=5 ∵c2=a2+b2 ⇒b2=64−25 ⇒25=9+b2⇒b2=16 Put a2=9 and b2=16 in Eq,(i),we get y29−x216=1 (iv)Since,foci (±5,0)lies on X-axis,as coordinate of Y is zero. Hence,equation of hyperbola wil be of the form x2a2−y2b2=1 ...(i) Where it is given that foci(±5,0)=(±c,0)and length of transverse axis is 2a=8 c=5 and 2a=8⇒a=4 ∵c2=a2+b2 ⇒25=16+b2⇒b2=9 Put a2=16 and b2=9 in Eq.(i),we get x216−y29=1 (v)Since,foci (0,±13)lies on Y-axis,as coordinate of X is zero. Hence,equation of hyperbola wil be of the form y2a2−x2b2=1 ...(i) Where it is given that foci(0,±13)=(0,±c)and conjugaete length 2b=24 c=13 and b=12 ∵c2=a2+b2 ⇒169=a2+144⇒a2=169−144 ⇒a2=25 Put a2=25 and b2=144 in Eq.(i),we get y225−x2144=1 (vi)Since,foci (±3√5,0)lies on X-axis,as y-coordinate is zero. Hence,equation of hyperbola wil be of the form y2a2−x2b2=1 ...(i) Where it is given that foci(±3√5,0)=(±c,0)and length oflatusrectum 2b2a=8 ∵b2=4a ∵c2=a2+b2 ∴(3√5)2=a2+(4a) ⇒9×5=a2+4a ⇒a2+4a−45=0 ⇒On spliting the middle term ⇒a2+9a−5a−45=0 ⇒a(a+9)−5(a+9)=0 ⇒(a−5)(a+9)=0 ⇒a−5=0,a+9=0 ⇒a=5,−9 ∵a can′t be negative. So,a=5⇒b2=4×5=20 Put a2=25 and b2=20 in Eq.(i),we get x225−y220=1 (vii)Since foci(±4,0)lies on X-axis,Hence,equation,of hyperbola will be of the form y2a2−x2b2=1 ...(i) Where it is given that foci(±4,0)=(±c,0)and length of latusrectum, $$\frac{2b^2}{a}=12\$$ ∴c=4 and b2=6a ...(ii) ∴c2=a2+b2 ⇒16=a2+6a ⇒a2+6a−16=0 Splitting the middle term a2+8a−2a−16=0 ⇒a(a(a+8)−2(a+8)=0 ⇒(a−2)(a+8)=0 ⇒a−2=0,a+8=0⇒a=2,−8 ∵a can′t be negative. So,a=2 ⇒b2=6a=6×2=12 Put a2=4 and b2=12 in Eq.(i),we get x24−y212=1 (viii)Let the equation of the hyperbola be −x2a2+y2b2=1 Then vertices=(0,±b)=(0,±6) ∴b=6 and e=54 ∵e=√1+a2b2⇒259=1+a236 ⇒25−99=a236⇒16=a24+48 So,the equation of hyperbola is, −x248+y236=1⇒y236−x248=1 ∵Foci=(0,±be)=(0,±53×6)=(0,±10) Since,foci(0,±√10)lies on Y-axis,as x-coordinate is zero. Hence,equation of hyperbola will be of the form y2a2−x2b2=1 ...(i) Where it is given that foci(0,±√10)=(0,±c)⇒c=√10 Also hyperbola passes through the point (1,3)i.e.,pont~will~satisfy~Eq.(i). ∴(3)2a2−(2)2b2=1 ⇒9a2−4b2=1 ...(ii) ∵c=√10 ⇒c2=√10 ⇒a2+b2=10 (∵c2=a2+b2) ⇒b2=10−a2 ...(iii) From Ep.(ii),9a2−410−a2=1 ⇒90−9a2−4aa2(10−a2)=1 ⇒90−13a2=10a2−a4 ⇒a4−23a2+90=0 ⇒a4−18a2−5a2+90=0 ⇒a2(a2−18)−5(a2−18)=0 ⇒a2=5 or a2=18 ⇒b2=10−5=5 If a2=18 ⇒b2=10−18=−8 Which in not possible. So,a2=b2=5 Put a2=b2=5 in Eq.(i)we get ⇒y25−x25=1⇒y2−x2=5 (x)Since foci are(0,±12),it follows that c=12 Length of the latus rectum=2b2a=36 or b2=18a Therefore c2=a2+b2,gives 144=a2+18a i.e.,a2+18a−144=0, So,a=−24,6 Sine a cannot be negative,we take a=6 and sob2=108 Therefore,the equation of the required hyperbola isy236−x2108=1, i.e.,3y2−x2=108

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